26
$\begingroup$

Sorry for the primitive question but when we inflate a rubber balloon and tie the end, its volume increases until its inner pressure equals atmospheric pressure.

But after that equality is obtained why does the air goes out when we pop the balloon? If there is pressure equality what causes the air flow?

Improve this question
$\endgroup$
4
  • 7
    $\begingroup$ The inner pressure does not equal the outer pressure. Rather, they are close enough that the difference can be ignored for most purposes (such as computing the buoyancy of the balloon). $\endgroup$
    – Hot Licks
    Commented Jun 9, 2019 at 2:59
  • 3
    $\begingroup$ Surely you mean a rubber balloon, not a plastic one? $\endgroup$
    – David Conrad
    Commented Jun 9, 2019 at 20:42
  • 2
    $\begingroup$ @DavidConrad true the elasticity of the balloon makes the air blow out. The analogy of a bin bag is best to understand this, as mentioned by Bilkokuya in the comment of the most voted answer. $\endgroup$
    – shabby
    Commented Jun 11, 2019 at 6:07
  • $\begingroup$ Can you explain why you believe that a pressurized balloon has the same pressure inside and out? In particular, suppose you have an uninflated balloon and you tie off the end: is the pressure inside the balloon in that case equal to, less than, or greater than the air pressure, in your conception of how the world works? $\endgroup$
    – Eric Lippert
    Commented Jun 11, 2019 at 20:37

6 Answers 6

Reset to default
60
$\begingroup$

For an inflated and tied balloon, the inner and outer pressures aren't equal. The inner pressure is higher by an amount $2 \gamma |H|$, where $\gamma$ is the inflated balloon's surface tension and $H$ is its mean curvature (which is $-1/R$ for a sphere). This is called the Young-Laplace equation.

After the balloon is untied and deflates, the pressures equalize and the surface tension becomes negligible.

Improve this answer
$\endgroup$
8
  • 3
    $\begingroup$ Not seen many balloons that are spherical... $\endgroup$
    – user207455
    Commented Jun 8, 2019 at 21:35
  • 7
    $\begingroup$ @SolarMike Weird - almost every balloon I've ever seen has been pretty close to spherical. $\endgroup$
    – tparker
    Commented Jun 9, 2019 at 13:56
  • 91
    $\begingroup$ They are certainly more spherical than any cow I've ever seen! $\endgroup$
    – Cort Ammon
    Commented Jun 9, 2019 at 15:16
  • 6
    $\begingroup$ For non-spherical balloons, the relationship between the mean curvature and the dimensions is different, but the main idea remains. $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jun 9, 2019 at 17:48
  • 7
    $\begingroup$ Easy way to prove this to yourself; get a bin bag and catch a bunch of air in it. Let the top of it open - notice that the air barely tries to escape, until you push the sides. The balloon is the same as this, but it's always pushing in from its own sides. $\endgroup$
    – Bilkokuya
    Commented Jun 10, 2019 at 10:05
29
$\begingroup$

But after that equality is obtained why does the air goes out when we penetrate the balloon? If there is pressure equality what causes the air flow?

You need to take into account that the elastic tension of the balloon skin pulls inwards. This makes the pressure in the balloon greater than its surroundings. Since there is a pressure difference the air blows out when you penetrate the skin defeating the elastic tension of the balloon skin.

Think about what happens when you blow up a balloon. At the end when the balloon gets taut it gets harder it to blow it up until it bursts. Clearly the outside pressure has not changed. The elastic tension of the balloon material has increased, like what happens when you stretch a rubber band just before it snaps.

Hope this helps.

Improve this answer
$\endgroup$
6
  • 10
    $\begingroup$ Actually, most balloons I know are much harder to blow up while they are small, especially at the point when we go from just unwrinkoiing the skin to actually expanding it - but this effect is due to material properties of the rubber and not relevant to the qualitative effect. $\endgroup$
    – Hagen von Eitzen
    Commented Jun 9, 2019 at 15:21
  • 2
    $\begingroup$ @BobD No, please see equation in tparker's answer. The balloon at the start has got smaller radius, hence greater curvature which requires higher pressure at the beginning. $\endgroup$
    – mpasko256
    Commented Jun 10, 2019 at 8:29
  • 5
    $\begingroup$ As a balloon expands, it is actually easier to blow up under certain differences in size. See the two-balloon experiment for an example. Until I read about this experiment on SE, I had never considered how balloon pressure changes with radius. physics.stackexchange.com/questions/317032/… $\endgroup$
    – JMac
    Commented Jun 10, 2019 at 13:27
  • 2
    $\begingroup$ @JMac At the beginning that's true. I was thinking of what happens at the end when the balloon gets very taut. I will edit to clarify. Thanks for your input. $\endgroup$
    – Bob D
    Commented Jun 10, 2019 at 13:37
  • 2
    $\begingroup$ @HagenvonEitzen If you watch professionals (either people selling normal balloons, or people making balloon animals), you will see that they start by stretching the rubber to warm it up and make it more pliable, which makes the initial inflation easier. If you just start blowing into a 'cold' balloon, you are trying to do that initial "warmup" with your lungs instead - and the balloon will also be stiffer and more prone to bursting too. $\endgroup$
    – Chronocidal
    Commented Jun 11, 2019 at 10:51
8
$\begingroup$

until its inner pressure equals to the atmospheric pressure

The inference that the balloon is not growing (or shrinking) because the pressure is the same is not correct.

The balloon is not growing because the effective force pushing the balloon out from inside is the same as the effective force pushing the balloon in from outside.

The force pushing outwards is indeed due to the pressure of the air inside the balloon.

But the force trying to collapse the balloon is the pressure of the air on the outside (atmospheric pressure) plus the elastics potential of the balloon trying to return to its original size and shape.

So, to counteract this additional force the pressure inside the balloon has to be higher than the air pressure outside the balloon.

Improve this answer
$\endgroup$
5
$\begingroup$

Blowing into a balloon is harder than just blowing into the air, because it takes higher air pressure to stretch the rubber. once the balloon is tied the stretched rubber continues to squeeze the air inside, so inner air pressure stays higher than outer air pressure. Untie the balloon and the stretched rubber will squeeze the air out until it shrinks to its normal un-stretched size. Sticking the inflated balloon with a needle will create a flaw in the stretched rubber causing it to split open and release the inner air pressure very fast, pop.

Improve this answer
$\endgroup$
0
$\begingroup$

I had the same doubt. But what i think as an answer is the pressure inside balloon is higher because not the atmosphere pressure is same as pressurein balloon, but because effective force is same. Let me explain Suppose a normal balloon, now you blow it what actuallyyou are doing is creating high pressure in balloon so maintain an equilibrium balloon expands , Now pressure is decreasing but ...It will not decrease upto atmospheric pressure but a little higher ,Why??!!

Because I am neglecting something, as balloon is expanding it is gaining an elastic potential or tension along with a little bit decrease in inner pressure ( as area increase pressure decrease ) so elastic potential is like a ...just an example u r squeezing balloon with your hand, I know it is very rough example still helpful! And yah remember the tension of balloon skin is on balloon itself not on the air inside ( this was my misconception that tension of balloon wall is applied on air inside) .

So there are kinda more net OUTSIDE force i.e atmosphere pressure plus tension of balloon on itself , so to balance this net force the INNER pressure has to be high comparative to "Alone" atmospheric pressure.

Hope it helps.

Any expert here please check if I am correct or not but this is my understanding , correct if I am wrong

Improve this answer
$\endgroup$
0
$\begingroup$

The pressure inside is greater until it balances with external pressure. The rubber is the only thing maintaining the balance. If it were any greater inside, it would continue to expand.

It takes very little pressure to lift large weight when applied to large surface areas. Since this minor pressure it applied to the entire inner surface it takes very little pressure to maintain.

Inflation takes pressure, pressure has to balance or inflation continues. It's not a steel belted tire. How much pressure does a thin rubber membrane take? How small a flaw destroys it?

I actually untied a helium balloon and let it go many years ago and it did not deflate. AS IT floated up we all could see into the balloon through the opening as it lifted, but did not deflate at all. It continued til out of sight.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.